Mathematics - Statistics Question with Solution | TestHub

Let the observations be $x_1, x_2, x_3, ..., x_8, 45, 50$,

Mean$=50$

$\Rightarrow \frac{x_1+x_2+x_3+...+x_8+45+50}{10}=50$

$\Rightarrow x_1+x_2+x_3+...+x_8=405 ....\left(\mathrm{i}\right)$

Hence, new mean

${\left(\overline{X}\right)}_{\mathrm{new}}=\frac{405+20+25}{10}=45$

Now,

$\mathrm{S}.\mathrm{D}=\sqrt{\frac{{\displaystyle \sum _{\mathrm{i}=1}^8{\mathrm{x}}_{\mathrm{i}}^2+{45}^2+{50}^2}}{10}-{\left(50\right)}^2}$

$\Rightarrow 12=\sqrt{\frac{{\displaystyle \sum _{\mathrm{i}=1}^8{\mathrm{x}}_{\mathrm{i}}^2+4525}}{10}-2500}$

$\Rightarrow \sum _{\mathrm{i}=1}^8{\mathrm{x}}_{\mathrm{i}}^2=21915$

Now,

${\left(\mathrm{Variance}\right)}_{\mathrm{new}}=\frac{{\displaystyle \sum _{\mathrm{i}=1}^8{\mathrm{x}}_{\mathrm{i}}^2+{20}^2+{25}^2}}{10}-{\left(45\right)}^2$

$\Rightarrow {\left(\mathrm{Variance}\right)}_{\mathrm{new}}=\frac{{\displaystyle 21915+{20}^2+{25}^2}}{10}-{\left(45\right)}^2$

$\Rightarrow {\left(\mathrm{Variance}\right)}_{\mathrm{new}}=2294-2025$

$\Rightarrow {\left(\mathrm{Variance}\right)}_{\mathrm{new}}=269$

Hence this is the correct answer.